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Given a string, consisting of a prefix and then "illion", convert this number into standard form.

For example:

"million" -> 10^6
"trillion" -> 10^12
"quattuordecillion" -> 10^45

The program needs to be able to handle input going up to Centillion, which is 10^303. A list of names and their standard form values can be found here - note that this gives values for every 10^3 increment up to 10^63, but then gives them in 10^30 increments, however the pattern is fairly straightforward.

The program needs to handle all 100 cases (even the ones not explicitly given by the website provided) - here are some examples of this:

"sexvigintillion" -> 10^81
"unnonagintillion" -> 10^276
"octotrigintillion" -> 10^117

Input can be given via STDIN, function argument or hard-coded as a string.

This is code-golf, so shortest code wins!

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  • \$\begingroup\$ What would 10^70 be? \$\endgroup\$
    – Scimonster
    Oct 28, 2014 at 18:56
  • 3
    \$\begingroup\$ 10^70 doesn't have a representation because 3 isn't a factor of 70 - but 10^69 would be sexvigintillion. 10^70 would be 10 sexvigintillion. \$\endgroup\$
    – James Williams
    Oct 28, 2014 at 19:02
  • \$\begingroup\$ Actually, doevigintillion = 10^69, and sexvigintillion = 10^81. \$\endgroup\$
    – Remy
    Oct 28, 2014 at 20:36
  • \$\begingroup\$ @Remy I would guess that you use the long scale (if that's right)? It sounds like this question uses the short scale. \$\endgroup\$
    – Cole Tobin
    Oct 29, 2014 at 2:59
  • \$\begingroup\$ @Cole Johnson: The question's provided list of names says vigintillion = 10^63, and shows that un- adds 3 to the power, doe- adds 6, sex- adds 18, etc. \$\endgroup\$
    – Remy
    Oct 29, 2014 at 3:15

5 Answers 5

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Python 2 (384 368 365 348 347 bytes)

def c(s):
 s=s[:-6].replace('int','');k=0;d=dict(un=1,doe=2,tre=3,quattuor=4,quin=5,sex=6,septen=7,octo=8,novem=9,b=3,tr=4,quadr=5,qu=6,sext=7,sept=8,oct=9,non=10,dec=11,vig=21,trig=31,quadrag=41,quinquag=51,sexag=61,septuag=71,octog=81,nonag=91,cent=101)
 for p in(s!='m')*list(d)*2:
    if s.endswith(p):s=s[:-len(p)];k+=3*d[p]
 return 10**(k or 6)

(The if line is indented with a single tab, and the rest with single spaces.)

Here c('million') == 10**6 has to be a special case because 'novem' also ends in 'm'.

Examples:

c('million') == 10**6
c('trillion') == 10**12
c('quattuordecillion') == 10**45
c('novemnonagintillion') == 10**300
c('centillion') == 10**303

Thanks to Falko for obfuscating it down to 350 bytes.

For practice I tried rewriting this as a one-liner using lambdas. It's 404 398 390 384 380 379 bytes:

c=lambda s:(lambda t=[s[:-5].replace('gint',''),0],**d:([t.__setslice__(0,2,[t[0][:-len(p)],t[1]+3*d[p]])for p in 2*list(d)if t[0].endswith(p)],10**t[1])[1])(un=1,doe=2,tre=3,quattuor=4,quin=5,sex=6,septen=7,octo=8,novem=9,mi=2,bi=3,tri=4,quadri=5,qui=6,sexti=7,septi=8,octi=9,noni=10,deci=11,vii=21,trii=31,quadrai=41,quinquai=51,sexai=61,septuai=71,octoi=81,nonai=91,centi=101)
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  • 2
    \$\begingroup\$ +1 for abusing OP's lack of specification of whether "10^x" should be printed or whether just returning the numerical value is sufficient. \$\endgroup\$
    – Ingo Bürk
    Oct 28, 2014 at 20:06
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    \$\begingroup\$ Thanks, although return'10^'+str(3*k) would only be 4 bytes more. \$\endgroup\$
    – Remy
    Oct 28, 2014 at 20:38
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    \$\begingroup\$ Since this is python 2, you could use a space indent for the first level, and tab for the second. You can also move both a and b into the function as keyword arguments. \$\endgroup\$
    – FryAmTheEggman
    Oct 28, 2014 at 20:44
  • 2
    \$\begingroup\$ 1000**k is shorter than 10**(3*k). Incrementing k by 3*d[p] is also equally short. \$\endgroup\$
    – xnor
    Oct 28, 2014 at 21:31
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    \$\begingroup\$ You can save a few characters by avoiding the early exit using if'm'==s:k=6;d=[] instead of a second lengthy return statement. \$\endgroup\$
    – Falko
    Oct 28, 2014 at 22:22
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JS (ES6), 292 270

Understands only the numbers written in the given list. The OP isn't clear about the others.

z=b=>{a="M0B0Tr0Quadr0Quint0Sext0Sept0Oct0Non0Dec0Undec0Doedec0Tredec0Quattuordec0Quindec0Sexdec0Septendec0Octodec0Novemdec0Vigint0Trigint0Quadragint0Quinquagint0Sexagint0Septuagint0Octogint0Nonagint0Cent".split(0);for(i in a)if(~b.indexOf(a[i]))return"10^"+(20>i?3*i+6:93+30*(i-20))}

Example:

z("Billion") // "10^9"
z("Centillion") // "10^303"
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  • \$\begingroup\$ You can remove the zeroes in the string and replace split(0) with match(/[A-Z][a-z]*/g) to use regexes to match each string. \$\endgroup\$
    – NinjaBearMonkey
    Oct 28, 2014 at 20:57
  • \$\begingroup\$ This only handles the "un, doe, tre, etc" prefixes for decillion. It should also handle cases like unvigintillion = 10^66 and novemnonagintillion = 10^300 \$\endgroup\$
    – Remy
    Oct 28, 2014 at 21:08
  • \$\begingroup\$ You can shorten this by using ES6 functions =>. \$\endgroup\$
    – soktinpk
    Oct 28, 2014 at 23:13
  • \$\begingroup\$ thanks for the tips. @Remy are you sure? the OP doesn't seem to ask that \$\endgroup\$
    – xem
    Oct 29, 2014 at 6:20
  • \$\begingroup\$ Seems clear to me that all multiples of 3 are required: "... this gives values for every 10^3 increment up to 10^63, but then gives them in 10^30 increments, however the pattern is fairly straightforward" in the OP. Also OP gives an example of "sexvigintillion" in a comment. \$\endgroup\$
    – feersum
    Oct 29, 2014 at 8:13
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C, 235

Handles all 100 cases. Program uses stdin and stdout.

Who needs regexes for camel-case splitting?

char*Z="UUUi+W<)E(<7-7-++*)('&%$,*$&%$",u[999]="\0MBRilDriPtiNiUnOeReTtUiXTeCtVeCiGRigRaUagInquiXaXsexPtuOgOoNaCeCeK1",s[99],*U=u+67;
main(n){
for(gets(s);*--U;)
*U<95?
*U|=32,
n+=!!strstr(s,U)*(*Z++-35),
*U=0:
3;puts(memset(u+68,48,3*n)-1);
}

Example

octoseptuagintillion
1000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
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3
  • 1
    \$\begingroup\$ This thing doesn't even look like C anymore... I'm amazed. \$\endgroup\$
    – Quentin
    Oct 29, 2014 at 12:29
  • \$\begingroup\$ Why the space (*U<95 ?) and all the newlines? \$\endgroup\$
    – tomsmeding
    Oct 29, 2014 at 14:25
  • \$\begingroup\$ @tomsmeding The space was an oversight. The newlines make the code "readable" and are not included in the count. \$\endgroup\$
    – feersum
    Oct 29, 2014 at 19:25
2
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Clojure, 381 377 bytes

(defn c[x](let[l{"M"6"B"9"Tr"12"Quadr"15"Quint"18"Sext"21"Sept"24"Oct"27"Non"30"Dec"33"Undec"36"Doedec"39"Tredec"42"Quattuordec"45"Quindec"48"Sexdec"51"Septendec"54"Octodec"57"Novemdec"60"Vigint"63"Trigint"93"Googol"100"Quadragint"123"Quinquagint"153"Sexagint"183"Septuagint"213"Octogint"243"Nonagint"273"Cent"303}v(l(clojure.string/replace x #"illion$" ""))](Math/pow 10 v)))

Example:

(c "Septuagintillion") ;; 1.0E213

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2
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Haskell, 204 bytes (+9 for formatted String)

import Data.List
x s=10^(f$[a|k<-tails s,i<-inits k,(b,a)<-zip["ce","ad","un","do","b","mi","vi","tr","at","ui","x","p","oc","no","ec","g"]$100:4:1:2:2:[1..],b==i])
f[]=3
f(x:11:r)=30*x+f r
f(x:r)=3*x+f r

In GHCi:

*Main> x "decillion"
1000000000000000000000000000000000

Replacing 10^( with "10^"++(show. adds another 9 bytes:

import Data.List
x s="10^"++(show.f$[a|k<-tails s,i<-inits k,(b,a)<-zip["ce","ad","un","do","b","mi","vi","tr","at","ui","x","p","oc","no","ec","g"]$100:4:1:2:2:[1..],b==i])
f[]=3
f(x:11:r)=30*x+f r
f(x:r)=3*x+f r

In GHCi:

*Main> x "decillion"
"10^33"

Edit: I had to correct for "quinquagintillion" which contains "qua".

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